Question

MAJOR HELP!
verify the identity

cotx/1+csc x = cscx -1/cotx

Answer 1

is this it?

\( \dfrac{\cot x}{1 + \csc x} = \dfrac{\csc x - 1}{\cot x} \)

Answer 2

yes it is it :)

Answer 3

\[\frac{ \cot x }{1+\csc x }*\frac{ \csc x-1 }{\csc x-1 }=\frac{ \cot x \left( \csc x-1 \right) }{ \csc ^{2}x-1 }\]
\[use \csc ^{2} x-\cot ^{2}x=1 \]
and get the solution

Answer 4

how do i get the solution?

Answer 5

its not multiplication they are equal to each other

Answer 6

\[\csc ^{2}x-1=\cot ^{2}x\]

Answer 7

\[=\frac{ \cot x \left( \csc x-1 \right) }{ \cot ^{2}x }=\frac{ \csc x-1 }{ \cot x }=R.H.S\]

Answer 8

Thanks so much :)

Answer 9

I have started with L.H.S and proved it equal to R.H.S

Answer 10

Here's a different approach. By working on both sides, you can show the two sides to be equal.

\( \dfrac{\cot x}{1 + \csc x} = \dfrac{\csc x - 1}{\cot x}\)

\( \dfrac{ \dfrac{\cos x}{\sin x}}{1 + \dfrac{1}{\sin x}} =
\dfrac{ \dfrac{1}{\sin x} - 1}{\dfrac{\cos x}{\sin x}} \)

\( \dfrac{\sin x}{\sin x} \dfrac{ \dfrac{\cos x}{\sin x}}{1 + \dfrac{1}{\sin x}} =
\dfrac{ \dfrac{1}{\sin x} - 1}{\dfrac{\cos x}{\sin x}} \dfrac{\sin x}{\sin x} \)

\( \dfrac{ \cos x}{\sin x + 1} =
\dfrac{ 1 -\sin x}{\cos x} \)

\( (\cos x)(\sin x + 1)\dfrac{ \cos x}{\sin x + 1} =
\dfrac{ 1 -\sin x}{\cos x} (\cos x)(1 +\sin x) \)

\(\cos^2 x = 1 - \sin^2 x\)

\( \cos^2 x = \cos^2 x\)